Let `a_1,a_2,a_3…., a_49` be in A.P . Such that `Sigma_(k=0)^(12) a_(4k+1)=416` and `a_9+a_(43)=66` .If `a_1^2+a_2^2 +…+ a_(17)` = 140 m then m is equal to

Options

A. 33
B. 66
C. 68
D. 34

Explanation

To solve the problem step by step, we will break down the information given and derive the necessary equations.

### Step 1: Understand the given information
We have an arithmetic progression (A.P.) defined by the terms \( a_1, a_2, a_3, \ldots, a_{49} \). The following information is provided:
1. \( \sum_{k=0}^{12} a_{4k+1} = 416 \)
2. \( a_9 + a_{43} = 66 \)

### Step 2: Express the sum in terms of \( a \) and \( d \)
The terms \( a_{4k+1} \) can be expressed as:
- \( a_1, a_5, a_9, a_{13}, a_{17}, a_{21}, a_{25}, a_{29}, a_{33}, a_{37}, a_{41}, a_{45}, a_{49} \)

Thus, we can write:
\[ \sum_{k=0}^{12} a_{4k+1} = a_1 + a_5 + a_9 + a_{13} + a_{17} + a_{21} + a_{25} + a_{29} + a_{33} + a_{37} + a_{41} + a_{45} + a_{49} \]
This can be simplified as:
\[ = 13 \cdot \frac{a_1 + a_{49}}{2} = 416 \]
where \( a_{49} = a_1 + 48d \).

### Step 3: Set up the equation
From the above expression, we have:
\[ \frac{13}{2} (a_1 + (a_1 + 48d)) = 416 \]
\[ \frac{13}{2} (2a_1 + 48d) = 416 \]
\[ 13(a_1 + 24d) = 416 \]
\[ a_1 + 24d = \frac{416}{13} = 32 \quad \text{(Equation 1)} \]

### Step 4: Use the second condition
From the second condition \( a_9 + a_{43} = 66 \):
\[ a_9 = a_1 + 8d, \quad a_{43} = a_1 + 42d \]
Thus:
\[ (a_1 + 8d) + (a_1 + 42d) = 66 \]
\[ 2a_1 + 50d = 66 \]
\[ a_1 + 25d = 33 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations
Now we have two equations:
1. \( a_1 + 24d = 32 \)
2. \( a_1 + 25d = 33 \)

Subtract Equation 1 from Equation 2:
\[ (a_1 + 25d) - (a_1 + 24d) = 33 - 32 \]
\[ d = 1 \]

Substituting \( d = 1 \) back into Equation 1:
\[ a_1 + 24(1) = 32 \]
\[ a_1 + 24 = 32 \]
\[ a_1 = 8 \]

### Step 6: Calculate \( a_1^2 + a_2^2 + \ldots + a_{17}^2 \)
The first 17 terms of the A.P. are:
\[ a_1 = 8, a_2 = 9, a_3 = 10, \ldots, a_{17} = 24 \]
We need to calculate:
\[ a_1^2 + a_2^2 + \ldots + a_{17}^2 = 8^2 + 9^2 + 10^2 + \ldots + 24^2 \]

### Step 7: Use the formula for the sum of squares
The sum of squares of the first \( n \) natural numbers is given by:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
We can calculate \( \sum_{k=1}^{24} k^2 \) and subtract \( \sum_{k=1}^{7} k^2 \):
\[ \sum_{k=1}^{24} k^2 = \frac{24 \cdot 25 \cdot 49}{6} = 4900 \]
\[ \sum_{k=1}^{7} k^2 = \frac{7 \cdot 8 \cdot 15}{6} = 140 \]
Thus:
\[ a_1^2 + a_2^2 + \ldots + a_{17}^2 = 4900 - 140 = 4760 \]

### Step 8: Set up the equation for \( m \)
We know that:
\[ a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m \]
So:
\[ 4760 = 140m \]
\[ m = \frac{4760}{140} = 34 \]

### Final Answer
Thus, the value of \( m \) is:
\[ \boxed{34} \]